Q. 33 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 31– 40, each matrix is non singular. Find the inverse of each matrix.
$\left[\begin{array}{cc}6& 5\\ 2& 2\end{array}\right]$

Step-by-Step Solution

Answer

The inverse of the matrix $\left[\begin{array}{cc}6& 5\\ 2& 2\end{array}\right]$ is $\left[\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}\right]$

Step 1. Given Information

The matrix $\left[\begin{array}{cc}6& 5\\ 2& 2\end{array}\right]$ is non singular.

Step 2. Form the matrix

$[A\overline{)I_2]=\left[\begin{array}{cc}6& 5\\ 2& 2\end{array}\right]\overline{)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}}$

Step 3. Transform equation into reduced row echelon form

$$\begin{gathered} R_1\to \frac{1}{6}{R}_1 \\ {R}_2\to R_2-2R_1=\left[\begin{array}{cc}1& \frac{5}{6}\\ 0& \frac{1}{3}\end{array}\right]\overline{)\left[\begin{array}{cc}\frac{1}{6}& 0\\ -\frac{1}{3}& 1\end{array}\right]} \\ {R}_2\to 3R_2=\left[\begin{array}{cc}1& \frac{5}{6}\\ 0& 1\end{array}\right]\overline{)\left[\begin{array}{cc}\frac{1}{6}& 0\\ -1& 3\end{array}\right]} \\ {R}_1\to R_1-\frac{5}{6}{R}_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\overline{)\left[\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}\right]} \end{gathered}$$

Step 4. Find the inverse

The inverse of A is $\left[\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}\right]$